Part 2 · The postulates

2.4 The Schrödinger equation

Source: Centini’s whiteboard of 9 Oct, p. 7–11; the plane wave, the free particle, the time-dependent equation and the evolution exercise from the 2024/25 boards #4, #5 and #20

Objectives

Understand which operator describes position, momentum and energy, and where each comes from.
Recipe: write the Hamiltonian of a problem and its time-independent Schrödinger equation; evolve a superposition in time.
Picture: a stationary state whose phase turns while $∣ ψ ∣^{2}$ stands still, and two levels that beat.
Theorem: $[\overset{x}{^}, \overset{p}{^}_{x}] = i ℏ$ ; every energy eigenstate times $e^{- i Et /ℏ}$ solves the time-dependent equation.
Vocabulary: de Broglie wavelength, Laplacian, free particle, stationary state.
Essential vocabulary: Hamiltonian.

Postulate 1 says each quantity is an operator. The 2024/25 recap lists the observables of the course: position, linear momentum, kinetic and potential energy (together, $\hat{H}$ ), orbital angular momentum (§4.2 Angular momentum) and spin (§4.1 Spin). Here are the first four.

Position and momentum

Position multiplies the wavefunction by the coordinate: $\overset{x}{^} ψ (\overset{r}{ˉ}) = x ψ (\overset{r}{ˉ}), \overset{y}{^} ψ (\overset{r}{ˉ}) = y ψ (\overset{r}{ˉ}), \overset{z}{^} ψ (\overset{r}{ˉ}) = z ψ (\overset{r}{ˉ}) .$

Momentum comes from de Broglie and Planck.

Definition 2.4.1De Broglie and Planck relations

A particle of momentum $p$ is a wave of wavelength $λ = \frac{h}{p}, so p = ℏ k,$ and a wave of angular frequency $ω$ carries energy $E = h ν = ℏ ω .$ $h = 6.63 \times 1 0^{- 34}$ J s is Planck’s constant and $ℏ = h /2 π$ .

So the plane wave of a free particle, written with its momentum and energy, is $ψ (x, t) = ψ_{0} e^{i k_{x} x} e^{- iω t} = ψ_{0} e^{i (p_{x} x - Et) /ℏ} .$ Its spatial part $e^{i p_{x} x /ℏ}$ is the eigenfunction of $- i ℏ \frac{d}{d x}$ with eigenvalue $p_{x}$ (Theorem 1.3.8). That reads off the operator: $\hat{\overset{p}{ˉ}} = - i ℏ \overset{ˉ}{\nabla} = - i ℏ (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}),$ the quantum counterpart of $\overset{p}{ˉ} = m \overset{v}{ˉ}$ . It is Hermitian (Theorem 1.4.10). In 3D its eigenfunctions are $e^{i \overset{p}{ˉ} \cdot \overset{r}{ˉ} /ℏ}$ , the plane waves of Definition 1.2.6 with $\overset{ˉ}{k} = \overset{p}{ˉ} /ℏ$ .

Example 2.4.2An electron of 1 eVbeyond the boards

$p = 2 m E = 2 \cdot 9.11 \times 1 0^{- 31} \cdot 1.60 \times 1 0^{- 19} = 5.40 \times 1 0^{- 25}$ kg m/s, so $λ = h / p = 1.23$ nm. ==Nanometre wavelengths at electron-volt energies==: that is why quantum effects show up in nanostructures.

Theorem 2.4.3Position and momentum do not commute

$[\overset{x}{^}, \overset{p}{^}_{x}] = i ℏ \hat{I}, [\overset{r}{^}_{i}, \overset{p}{^}_{j}] = i ℏ δ_{ij} \hat{I} .$

Proof

Apply the commutator to a test function $f (x)$ (Recipe 2.3.3) and write $g = f^{'}$ : $[\overset{x}{^}, \overset{p}{^}_{x}] f = \overset{x}{^} \overset{p}{^}_{x} f - \overset{p}{^}_{x} (\overset{x}{^} f) = - i ℏ x f^{'} + i ℏ (x f)^{'} = - i ℏ x g + i ℏ (f + x g) = i ℏ f .$ The product rule gives $(x f)^{'} = f + x f^{'}$ , and the $xg$ terms cancel. This holds for every $f$ , and $\hat{I} f = f$ . Along different directions, $\frac{\partial}{\partial x}$ treats $y$ as a constant, so $[\overset{y}{^}, \overset{p}{^}_{x}] = 0$ .

==Position and momentum along the same axis are incompatible==: no state has both sharp. This is the operator form of the bandwidth relation, and its quantitative form is §2.5 The uncertainty principle.

The Hamiltonian

Classically, energy is kinetic plus potential, $E = E_{k} + U$ . The potential energy $U (\overset{r}{ˉ})$ is a function of position, so its operator multiplies by $U$ , like $\overset{x}{^}$ . The kinetic energy, written with momentum, is $E_{k} = \frac{1}{2} m v^{2} = \frac{p ^{2}}{2 m}$ . Replace $\overset{p}{ˉ}$ by its operator: $\hat{E}_{k} = \frac{1}{2 m} \hat{\overset{p}{ˉ}} \cdot \hat{\overset{p}{ˉ}} = \frac{( - i ℏ ) ^{2}}{2 m} \overset{ˉ}{\nabla} \cdot \overset{ˉ}{\nabla} = - \frac{ℏ ^{2}}{2 m} \nabla^{2},$ with the Laplacian $\nabla^{2} = \frac{\partial ^{2}}{\partial x ^{2}} + \frac{\partial ^{2}}{\partial y ^{2}} + \frac{\partial ^{2}}{\partial z ^{2}}$ . (This year’s board drops the square in the last step and writes $- ℏ \nabla^{2}$ ; the 2024/25 board has $ℏ^{2}$ .)

Definition 2.4.4HamiltonianEssential

The energy operator is the Hamiltonian $\hat{H} = \hat{E}_{k} + \hat{U} = - \frac{ℏ ^{2}}{2 m} \nabla^{2} + U (\overset{r}{ˉ}) .$

$\nabla^{2} ψ$ measures how sharply $ψ$ curves. ==More curvature means shorter wavelength, more momentum, more kinetic energy==: squeezing a wavefunction into a small region costs energy (beyond the boards).

The time-independent equation

Theorem 2.4.5Time-independent Schrödinger equation

The states of definite energy are the eigenstates of $\hat{H}$ , $\hat{H} u_{E} = E u_{E}$ : $[- \frac{ℏ ^{2}}{2 m} \nabla^{2} + U (\overset{r}{ˉ})] u_{E} (\overset{r}{ˉ}) = E u_{E} (\overset{r}{ˉ}) .$

$u_{E}$ is an energy eigenfunction, $E$ its energy, what a measurement of energy returns in that state. $U$ is where each problem differs. The 2024/25 recap lists the one-dimensional ones the course solves: the free particle (below), the infinite well, the finite well, the barrier and the harmonic oscillator.

Recipe 2.4.6Setting up a problem

Write $U$ . In 1D, $\nabla^{2} \to \frac{d ^{2}}{d x ^{2}}$ : $- \frac{ℏ ^{2}}{2 m} \frac{d ^{2} u}{d x ^{2}} + U (x) u = E u .$
Where $U$ is infinite, $u = 0$ .
Solve region by region, and impose what joins the regions and what keeps $u$ normalisable (§3.1 One-dimensional problems).
The energies allowed by those conditions are the eigenvalues.

The free particle

With $U = 0$ in 1D, $\hat{H} = - \frac{ℏ ^{2}}{2 m} \frac{d ^{2}}{d x ^{2}}$ . Try the momentum eigenfunctions $u_{p} = A e^{i p_{x} x /ℏ}$ : each derivative brings down $\frac{i p _{x}}{ℏ}$ , $i \cdot i = - 1$ cancels the minus, and $\hat{H} u_{p} = \frac{p _{x}^{2}}{2 m} u_{p} .$ ==The energy is the classical kinetic energy, and any $p_{x}$ is allowed==: the free particle’s energies are not quantised. It works because here $\hat{H} = \overset{p}{^}_{x}^{2} /2 m$ commutes with $\overset{p}{^}_{x}$ , so the two share eigenfunctions (Theorem 2.3.2). Each energy has two, $p_{x} = \pm 2 m E$ : a wave travelling right and one travelling left (beyond the boards).

The catch is normalisation, $\int ∣ A ∣^{2} d x = \infty$ . The board puts the particle in a periodic box and lets the box grow, which gives $u_{p} = e^{i p_{x} x /ℏ} / 2 π ℏ$ (Key observation 1.5.6).

The time-dependent equation

Theorem 2.4.7Time-dependent Schrödinger equation2024/25 board #5

The state changes in time as $\hat{H} ψ (\overset{r}{ˉ}, t) = i ℏ \frac{\partial}{\partial t} ψ (\overset{r}{ˉ}, t) .$

It is a postulate, not derived. $i ℏ \frac{\partial}{\partial t}$ does to the time part what $\overset{p}{^}_{x} = - i ℏ \frac{\partial}{\partial x}$ does to the spatial part: $i ℏ \frac{\partial}{\partial t} e^{- i Et /ℏ} = E e^{- i Et /ℏ}$ , reading the energy off the rate at which the phase turns (beyond the boards). On the free plane wave $e^{i (k x - ω t)}$ , the two sides agree when $ℏ ω = ℏ^{2} k^{2} /2 m$ : $E = p^{2} /2 m$ again.

Key observation 2.4.8Stationary statesbeyond the boards

If $\hat{H} u_{E} = E u_{E}$ , then $ψ (\overset{r}{ˉ}, t) = u_{E} (\overset{r}{ˉ}) e^{- i Et /ℏ}$ solves the time-dependent equation. $∣ ψ ∣^{2} = ∣ u_{E} ∣^{2}$ does not change in time: a stationary state. ==Solve the time-independent equation, then multiply each solution by its phase.==

Proofbeyond the boards

$\hat{H}$ does not touch $t$ : $\hat{H} ψ = e^{- i Et /ℏ} \hat{H} u_{E} = E u_{E} e^{- i Et /ℏ}$ . And $i ℏ \frac{\partial ψ}{\partial t} = i ℏ (- \frac{i E}{ℏ}) u_{E} e^{- i Et /ℏ} = E u_{E} e^{- i Et /ℏ}$ . Equal.

Superpositions. The equation is linear, so a sum of stationary states evolves term by term, each with its own phase: $∣ ψ (0)⟩ = n \sum c_{n} ∣ ϕ_{n} ⟩ ⟹ ∣ ψ (t)⟩ = n \sum c_{n} e^{- i E_{n} t /ℏ} ∣ ϕ_{n} ⟩ .$ The probability of each energy stays $∣ c_{n} ∣^{2}$ (2024/25 board #20), so $⟨ \hat{H} ⟩$ is constant. The relative phases do change, so other quantities, $⟨ \overset{x}{^} ⟩$ say, can move.

Loading the demo…

Figure 2.4.1 An electron in a 1 nm infinite well (§3.2 The infinite well). With one level only, the

real

and

imaginary

parts of

ψ

swap back and forth as the phase turns, but

∣ ψ ∣^{2}

stands still. Raise

∣ c_{2} ∣^{2}

: two levels beat, and

∣ ψ ∣^{2}

and

⟨ x ⟩

slosh across the well at

ω = (E_{2} - E_{1}) /ℏ

. Time is in units of

h / E_{1}

Exercises

Exercise 2.4.1Normalisation and time evolution2024/25 board #20

A system starts in $∣ ψ (0)⟩ = \frac{A}{12} ∣ ϕ_{1} ⟩ + \frac{1}{6} ∣ ϕ_{2} ⟩ + \frac{2}{12} ∣ ϕ_{3} ⟩ + \frac{1}{2} ∣ ϕ_{4} ⟩$ , with orthonormal energy eigenstates. (a) Find $A$ . (b) Write $∣ ψ (t)⟩$ . © Probability of finding the system in $∣ ϕ_{2} ⟩$ at time $t$ ?

Solution

(a) $\frac{A ^{2}}{12} + \frac{1}{6} + \frac{4}{12} + \frac{1}{4} = 1$ , so $\frac{A ^{2}}{12} = \frac{3}{12}$ , $A = 3$ (any $3 e^{i α}$ works too). (b) $∣ ψ (t)⟩ = \frac{3}{12} e^{- i E_{1} t /ℏ} ∣ ϕ_{1} ⟩ + \frac{1}{6} e^{- i E_{2} t /ℏ} ∣ ϕ_{2} ⟩ + \frac{2}{12} e^{- i E_{3} t /ℏ} ∣ ϕ_{3} ⟩ + \frac{1}{2} e^{- i E_{4} t /ℏ} ∣ ϕ_{4} ⟩$ . © $\frac{1}{6} e^{- i E_{2} t /ℏ}^{2} = \frac{1}{6}$ , at every $t$ .

Exercise 2.4.2Theory 1(i), sample3

A particle is in a one-dimensional infinite square well, $V = 0$ for $- L /2 \leq x \leq L /2$ and $V = \infty$ elsewhere. Write the Hamiltonian and the time-independent Schrödinger equation.

Solution

$\hat{H} = - \frac{ℏ ^{2}}{2 m} \frac{d ^{2}}{d x ^{2}} + V (x)$ , and $- \frac{ℏ ^{2}}{2 m} u^{''} + V (x) u = E u$ . Inside, $V = 0$ : the free-particle equation. Outside, $V u$ stays finite only if $u = 0$ , and $u$ is continuous, so ==the walls impose $u (\pm L /2) = 0$ ==. The rest is §3.2 The infinite well.

Exercise 2.4.3Sample questions, Q2 (v)–(vi)

(v) The de Broglie wavelength of a 60 kg runner at 20 km/h; what does it tell you? (vi) Evaluate $[\overset{x}{^}, \overset{p}{^}]$ ; what does it mean?

Solution

(v) $v = 5.56$ m/s, $p = 333$ kg m/s, $λ = h / p = 2.0 \times 1 0^{- 36}$ m, some $1 0^{- 21}$ times a proton: no obstacle could show the runner diffract. Classical mechanics is the limit. (vi) $i ℏ \hat{I}$ (Theorem 2.4.3): position and momentum along one axis are incompatible and cannot both be sharp.

Exercise 2.4.4beyond the boards

$ψ (x, t) = \frac{1}{2} [u_{1} e^{- i E_{1} t /ℏ} + u_{2} e^{- i E_{2} t /ℏ}]$ with $u_{1}$ , $u_{2}$ real and orthonormal. Is $∣ ψ ∣^{2}$ stationary?

Solution

$∣ ψ ∣^{2} = \frac{1}{2} [u_{1}^{2} + u_{2}^{2} + 2 u_{1} u_{2} cos \frac{( E _{2} - E _{1} ) t}{ℏ}]$ . No: it sloshes at $(E_{2} - E_{1}) /ℏ$ , as in the figure, though each energy keeps probability $\frac{1}{2}$ . Only a single energy eigenstate is stationary.

Exercise 2.4.5beyond the boards

Check that $\hat{H} = \overset{p}{^}_{x}^{2} /2 m$ and $\overset{p}{^}_{x}$ commute, and say why the free particle’s $u_{E}$ can be chosen as momentum eigenstates.

Solution

$[\overset{p}{^}_{x}^{2}, \overset{p}{^}_{x}] = \overset{p}{^}_{x}^{3} - \overset{p}{^}_{x}^{3} = 0$ . Commuting observables have a common eigenbasis (The Compatible Observables Theorem), and $e^{\pm i p_{x} x /ℏ}$ is that basis: each is an energy eigenstate with $E = p_{x}^{2} /2 m$ .